Minitab® – One Sample Mean t Test Using Summarized Data
Here we are testing \(H_{a}\colon\mu\neq72\) and are given \(n=35\), \(\bar{x}=76.8\), and \(s=11.62\).
We do not know the shape of the population, however the sample size is large (\(n \ge 30\)) therefore we can conduct a one sample mean \(t\) test.
To perform a one sample mean \(t\) test in Minitab using raw data:
- In Minitab, select Stat > Basic Statistics > 1-sample t
- Select Summarized data from the dropdown
- Enter 35 for the sample size, 76.8 for the sample mean and 11.62 for the standard deviation.
- Check the box Perform a hypothesis test
- For the Hypothesized mean enter 72
- Select Options
- Use the default Alternative hypothesis of Mean ≠ hypothesized value
- Use the default Confidence level of 95
- Click OK and OK
This should result in the following output:
Descriptive Statistics
| N | Mean | StDev | SE Mean | 95% CI for \(\mu\) |
|---|---|---|---|---|
| 35 | 76.80 | 11.62 | 1.96 | (72.81, 80.79) |
| Null hypothesis | H0: \(\mu\) = 72 |
|---|---|
| Alternative hypothesis | H1: \(\mu\) ≠ 72 |
| T-Value | P-Value |
|---|---|
| 2.44 | 0.0199 |
We could summarize these results using the five step hypothesis testing procedure:
1. Check assumptions and write hypotheses
The shape of the population distribution is unknown, however with \(n \ge 30\) we can perform a one sample mean t test.
\(H_0\colon \mu = 72\)
\(H_a\colon \mu \ne 72\)
2. Calculate the test statistic
\(t (34) = 2.44\)
3. Determine the p-value
\(p = 0.0199\)
4. Make a decision
\(p \le \alpha\), reject the null hypothesis
5. State a "real world" conclusion
There is evidence that the population mean is different from 72.